Question

find the values of y for which the distance between the points P ( 2 , -3 ) and Q ( 10 , y ) is 10 units​

Answer 1

Step-by-step explanation:

A.T.Q.

10 = √(x2 - x1)^2 + (y2 - y1)^2

10 = √(10-2)^2 + (y+3)^2

10 = √ 64 + y^2 + 6y + 9

Squaring both sides

100 = 73 + y^2 + 6y

y^2 + 6y - 27 = 0

y^2 + 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

(y - 3)(y + 9) = 0

Therefore, the 2 possible values of y are 3 and -9.

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